Statistical Thermodynamics and Rate Theories/Eyring Transition State Theory

Figure 1: Reaction coordinate diagram for the bimolecular nucleophilic substitution (S_N2) reaction between bromomethane and the hydroxide anion

Chemical kinetics describes the rates of change of the concentration of species through chemical reactions.

Rate=$k[A{]}^{v_a}[B{]}^{v_b}$

The rate constant, k, is depends on the chemical species undergoing reaction and the reaction conditions (i.e, temperature).

Rate constants of elementary gas phase chemical reactions can be estimated by statistical thermodynamics using a rate theory. One of the simplest and more widely used theories is Eyring conventional transition state theory.

In Eyring conventional transition state theory, the rate constant of a chemical reaction can be expressed in terms of the partition functions of the transition state and the reactants. $\mathrm{\Delta }{E}^{‡}$ is the barrier height of the reaction.

$k=\frac{k_{B}T}{h}\frac{q_{TS}/V}{\prod _i^{reactants}{(q_i/V)}^{v_i}}\text{exp}\left(\frac{-\mathrm{\Delta }{E}^{‡}}{k_{B}T}\right)$

Assumptions of Transition State Theory

Transition state theory relies on three key assumptions in its derivation.

1. Reactants are in constant equilibrium with the transition state structure.
2. The energy of the particles follow a Boltzmann distribution.
3. Once reactants become the transition state, the transition state structure does not collapse back to the reactants.

Given a reaction where reactants A and B are converted to a transition state C:

$\mathrm{A}+\mathrm{B}\stackrel{-\rightharpoonup }{\leftharpoondown -}[\mathrm{C}]{\phantom{A}}^{‡}$

The equilibrium constant, K^‡, can be written in terms of the concentrations of the reactants and the transition state.

$K^{‡}=\frac{[\mathrm{C}]{\phantom{A}}^{‡}}{[\mathrm{A}][\mathrm{B}]}$

This reaction exists as an equilibrium because we assume that not every collision results in the formation of the transition state. Rewriting the concentrations in terms of the partition functions yields the equation below.

$k=\frac{k_{B}T}{h}\frac{q_{TS}/V}{\prod _i^{reactants}{(q_i/V)}^{v_i}}\text{exp}\left(\frac{-\mathrm{\Delta }{E}^{‡}}{k_{B}T}\right)$

Thermodynamic Representation of Eyring Equation

The Eyring equation can also be expressed in terms thermodynamic parameters. The standard free energy, ΔG^‡, can be written in terms of the equilibrium constant.

$\mathrm{\Delta }{G}^{‡}=-RT\ln (K^{‡})$

$K^{‡}=e^{\frac{-\mathrm{\Delta }{G}^{‡}}{RT}}$

Using the equation for standard Gibb's energy, the Eyring equation can be written as:

$k=\frac{k_{B}T}{h}{e}^{\frac{-(\mathrm{\Delta }{H}^{‡}-T\mathrm{\Delta }{S}^{‡})}{RT}}$

Here, $\mathrm{\Delta }{H}^{‡}$is the enthalpy of activation and represents the enthalpy or energetic stability of the transition state structure; it is the energy needed to break the bonds of the reactants and form the transition state. Similar to standard enthalpy, a negative value represents a more stable energetic structure and vice versa. $\mathrm{\Delta }{S}^{‡}$is the entropy of activation, which represents the entropy, or disorder of the transition state and is the number of accessible states of transition state compared to the reactants. Likewise, a positive entropy represents a more disordered system while a negative entropy corresponds to a more ordered system.

Using the thermodynamical representation of the Eyring equation allows for the prediction of the type of mechanism a reaction can take. An associative mechanism involves a molecule bonding to another to form the transition state, while a dissociative mechanism involves the breaking of a molecule to form the transition state. In general, an associative mechanism occurs when the entropy of the transition state is lower than the reactants and thus the transition state is more ordered; this means change in entropy is negative. the result of this is that the entropy is lower and the structure is more ordered and has less accessible states. Contrary, a dissociative mechanism occurs when the entropy of the transition state is positive or higher for the transition state relative to the reactants. Thus, the transition state is less ordered than the initial reactants.

Partition functions of the transition state

TST requires the calculation of the partition functions for the transition state structure.

$q_{TS}=q_{trans,TS}\times q_{rot,TS}\times q_{vib,TS}^{\text{'}}\times q_{el,TS}$

The translational and rotational partition functions of the transition state have the same form as those used for the reactant species. The vibrational partition function differs slightly because the vibration that corresponds to crossing the barrier is removed from this term (it "becomes" the $k_{B}T/h$ term. As result, there is one fewer term in the vibrational partition function. For non-linear transition states, there will be $3N_{atom}-5$ terms. For linear transition states, there will be $3N_{atom}-4$ terms.

$q_{vib,TS}^{\text{'}}=\prod _i^{n_{vib}-1}\frac{1}{1-\exp \left(\frac{-h\nu }{k_{B}T}\right)}$

Calculation of a rate constant using this theory requires that all the structural and mechanical properties of the transition state be known. This is a challenge because it is very difficult to use spectroscopy to determine the structural properties of a transition state, which typically only exists for 10-200 fs. Instead, computational chemistry is commonly used to calculate the properties of transition states.

Example

Calculate the rate constant for the reaction of H + H₂ at 300 K.

H + H₂ → H₂ + H

To calculate the rate constant the following are required, for both the reactants and the transition state:

• the masses of isotopes
• vibrational frequencies (q_vib)
• moments of inertia (q_rot)
• electronic degeneracies'
• reaction barrier height

Species	Vibration	$\nu$(cm-1)
H2	H-H stretch	4401.1272
H-H-H‡	H-H-H bend (degeneracy of 2)	875.62
H-H-H‡	H-H-H symmetric stretch	2050.67

R = 0.743 Å for the reactant R = 0.931 Å for the transition state $\mathrm{\Delta }E$^‡ = 38.61 kJ mol^-1

translational partition functions:

$q_{trans}(V,T)={\left(\frac{2\pi k_{B}Tm}{h^2}\right)}^{3/2}$

for H₂: $q_{trans}(V,T)={\left(\frac{2\pi (1.381\times 10^{-23}J{K}^{-1})(300K)(2\times 1.6735575\times 10^{-27}kg)}{(6.626\times 10^{-34}Js{)}^2}\right)}^{3/2}$ $q_{trans}(V,T)=2.795\times 10^{29}{m}^{-3}$

For H-H-H:$q_{trans}(V,T)={\left(\frac{2\pi (1.381\times 10^{-23}J{K}^{-1})(300K)(3\times 1.6735575\times 10^{-27}kg)}{(6.626\times 10^{-34}Js{)}^2}\right)}^{3/2}$ $q_{trans}(V,T)=5.135\times 10^{30}{m}^{-3}$

rotational partition functions:

for H₂ : $q_{rot}(T)=\left(\frac{2k_{B}T\mu r_e^2}{\sigma {\hslash }^2}\right)$

         $q_{rot}(T)=\left(\frac{(2)(1.381\times 10^{-23}J{K}^{-1})(300K)(0.5)(1.674\times 10^{-27}kg)(7.43\times 10^{-11}m{)}^2}{2(1.058\times 10^{-34}Js{)}^2}\right)$
         $q_{rot}(T)=1.71$

For H-H-H: $q_{rot}(T)=\left(\frac{2k_{B}T(2m_A{R}^2)}{\sigma {\hslash }^2}\right)$

          $q_{rot}(T)=\left(\frac{2(1.381\times 10^{-27}J{K}^{-1}(300K)(2)(1.674\times 10^{-27}kg)(9.31\times 10^{-11}m{)}^2}{2(1.058\times 10^{-34}Js{)}^2}\right)$
          $q_{rot}(T)=10.7$

vibrational partition functions:

for H₂:$q_{vib}(T)=\left(\frac{1}{1-\exp \frac{-hc{\nu }_1}{k_{b}T}}\right)\left(\frac{1}{1-\exp \frac{-hc{\nu }_2}{k_{B}T}}\right)\left(\frac{1}{1-\exp \frac{-hc{\nu }_3}{k_{B}T}}\right)$

                  $q_{vib}(T)=\left(\frac{1}{1-\exp \frac{-hc(875.62c{m}^{-1})}{k_{B}T}}\right)\left(\frac{1}{1-\exp \frac{-hc(875.62c{m}^{-1})}{k_{B}T}}\right)\left(\frac{1}{1-\exp \frac{-hc(2043.95c{m}^{-1})}{k_{B}T}}\right)$
                  $q_{vib}(T)=(1.015)(1.015)(1.000)$
                  $q_{vib}(T)=1.030$

Rate constant k:

$k=\left(\frac{k_{B}T}{h}\right)\left(\frac{q_{trans,TS}/V}{(q_{trans,H}/V)(q_{trans,H_2}/V)}\right)\left(\frac{q_{rot,TS}}{q_{rot,H_2}/V}\right)\left(\frac{q{\text{'}}_{vib,H_2}}{q_{vib,H_2}}\right)\left(\frac{g_{TS}}{g_H{g}_{H_2}}\right)e^{-\mathrm{\Delta }{E}^{‡}/k_{B}T}$

$k=\left(\frac{(1.381\times 10^{-23}J{K}^{-1})(300K)}{6.626\times 10^{-34}Js}\right)\left(\frac{5.135\times 10^{30}{m}^{-3}}{(9.883\times 10^{29}{m}^{-3})(2.795\times 10^{30}{m}^{-3})}\right)\left(\frac{10.7}{1.71}\right)\left(\frac{1.030}{1.00}\right)\left(\frac{2}{2\times 1}\right)e^{-38610Jmo{l}^{-1}/6.022\times 10^{23}mo{l}^{-1}/1.307\times 10^{-23}J{K}^{-1}\times 300K}$ $k=(6.251\times 10^{12}{s}^{-1})(1.8588\times 10^{-30}{m}^3)(6.25)(1.03)(1)(1.894\times 10^{-7})$

$k=1.42\times 10^{-23}{m}^3{s}^{-1}$

References

Atkins, P., de Paula, J., Elements of Physical Chemistry, 5th ed.; Oxford University Press, 2009.